PYTHON · NUMPY

What recalibration is worth

A cheap node feeds an expensive decision. What happens to its calibration interval?

Price a calibration interval against the cost of the decisions it protects, and find the interval that minimises total cost.

01Challenge

A farm manager refuses a quarterly calibration contract because it costs more than the annual one. Work out whether they are right, using the cost of the decisions the readings feed.

Predict firstA cheap node feeds an expensive decision. What happens to its calibration interval?

02Model

Calibration is usually argued as a quality obligation, which is why it loses to a budget. It is better argued as an expected cost, because then it competes on the same axis as everything else. Two costs move in opposite directions. Calibrating more often costs more in labour and lost service time. Calibrating less often costs more in decisions made on readings that had drifted past the difference they needed to resolve. Their sum has a minimum, and that minimum is the defensible interval. What makes this worth computing rather than asserting is that the answer depends on the value of the decision, not on the price of the sensor. A cheap node feeding an expensive decision earns frequent calibration, and an expensive node feeding a decision nothing turns on does not.

03Guided practice

Write both costs as functions of the interval in months. Labour is the number of visits per year times the cost of a visit. Decision loss is the probability a reading has drifted past the threshold, times how often the decision is made, times what a wrong one costs. Add them and search the interval that gives the smallest total.

04Feedback

PASS WHEN You produced an interval from the value of the decision rather than from the price of the sensor, which is the only way the argument survives a budget meeting.

If your run fails, check:

labour should count visits per year, which is twelve divided by the interval, times VISIT_COST.
decision_loss needs the fraction of the interval spent past THRESHOLD. Drift reaches the threshold at THRESHOLD/DRIFT_PER_YEAR years; anything after that is exposed.
If the total never has a minimum, check that labour falls and loss rises with the interval. One of them is probably the wrong way round.

05Retrieve & space

Calibration frequency is an expected-cost problem, not a quality obligation.
The two costs move in opposite directions and their sum has a minimum.
The interval is set by the value of the decision, not by the price of the sensor.

06Mastery & project

State the condition under which annual calibration is the correct choice, in terms of the quantities in this bench.

Price the calibration interval for one real system: find the visit cost, the decision frequency and the cost of a wrong decision. Where a number is unavailable, say so and give the range over which the answer would change.

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PYTHON · NUMPY

What recalibration is worth

Both cost curves are yours to write. Find the interval that minimises their sum, then test whether the answer survives a tenfold change in the value of the decision.

You read

the arrays and values already in scope

You change

the code you write in each cell

Fixed

the dataset and the checks that grade you